memory layout of inherited class

Question

I'm keen to know exactly how the classes will be arranged in memory esp. with inheritance and virtual functions.

I know that this is not defined by the c++ language standard. However, is there any easy way to find out how your specific compiler will implement these say by writing some test code?

EDIT:- Using some of the answers below :-

#include <iostream>

using namespace std;

class A {
  public:
    int a;
    virtual void func() {}
};

class B : public A {
  public:
    int b;
    virtual void func() {}
};

class C {
  public:
    int c;
    virtual void func() {}
};

class D : public A, public C {
  public:
    int d;
    virtual void func() {}
};

class E : public C, public A {
  public:
    int e;
    virtual void func() {}
};

class F : public A {
  public:
    int f;
    virtual void func() {}
};

class G : public B, public F {
  public:
    int g;
    virtual void func() {}
};

int main() {
  A a; B b; C c; D d; E e; F f; G g;
  cout<<"A: "<<(size_t)&a.a-(size_t)&a<<"\n";
  cout<<"B: "<<(size_t)&b.a-(size_t)&b<<" "<<(size_t)&b.b-(size_t)&b<<"\n";
  cout<<"C: "<<(size_t)&c.c-(size_t)&c<<"\n";
  cout<<"D: "<<(size_t)&d.a-(size_t)&d<<" "<<(size_t)&d.c-(size_t)&d<<" "<<(size_t)&d.d-    (size_t)&d<<"\n";
  cout<<"E: "<<(size_t)&e.a-(size_t)&e<<" "<<(size_t)&e.c-(size_t)&e<<" "<<(size_t)&e.e-    (size_t)&e<<"\n";
  cout<<"F: "<<(size_t)&f.a-(size_t)&f<<" "<<(size_t)&f.f-(size_t)&f<<"\n";
  cout<<"G: "<<(size_t)&g.B::a-(size_t)&g<<" "<<(size_t)&g.F::a-(size_t)&g<<" "    <<(size_t)&g.b-(size_t)&g<<" "<<(size_t)&g.f-(size_t)&g<<" "<<(size_t)&g.g-(size_t)&g<<"\n";
}

And the output is :-

A: 8
B: 8 12
C: 8
D: 8 24 28
E: 24 8 28
F: 8 12
G: 8 24 12 28 32

So all classes have got a v-ptr at loc 0 of size 8. D has another v-ptr at location 16. Similarly for E. G also seems to have a v-ptr at 16 although from my (limited) understanding I would have guessed it to have more.

Answer 1

As long as you stick to single inheritance, the subobjects are typically layed out in the order they are declared. A pointer is prepended to they type information at the front which is e.g. used for dynamic dispatch. Once multiple inheritance is incolved things become more complex, especially when virtual inheritance is involved.

To find precise information at least for one ABI flavour you can look for the Itanium ABI. This documents all these details. It is used as the C++ ABI at least on some Linux platforms (i.e. there multiple compilers can produce object files linked into one executable).

To determine the layout just print the addresses of subobjects of a give object. That said, unless you happen to implement a compiler it typically doesn't matter. The only real use of the object layout I doubd is arranging members to minimize padding.

Answer 2

The best way is probably writing a few simple test cases and then compile and debug them in assembler (all optimization off): running one instruction at a time you'll see where everything fits.

At least that's the way I learned it.

And if you find any case particularly challenging, post in in SO!

Answer 3

Visual Studio atleast has a hidden compiler option /d1reportSingleClassLayout (starting at ~32:00).

Usage: /d1reportSingleClassLayoutCLASSNAME where there shall be no whitespace between the compiler switch and CLASSNAME (obviously replace this with the name of the class you'e interested in).

Answer 4

Create an object of class, cast pointer to it to your machine's word, use sizeof to find the size of the object, and examine the memory at the location. Something like this:

#include <iostream>

class A
{
 public:
  unsigned long long int mData;
  A() :
   mData( 1 )
  {
  }      
  virtual ~A()
  {
  }
};
class B : public A
{
 public:
  unsigned long long int mData1;
  B() :
   A(), mData1( 2 )
  {
  }
};

int main( void )
{
 B lB;

 unsigned long long int * pB = ( unsigned long long int * )( &lB );

 for( int i = 0; i < sizeof(B) / 8; i++ )
 {
  std::cout << *( pB + i ) << std::endl;
 }

 return ( 0 );
}


Program output (MSVC++ x86-64):

5358814688 // vptr
1          // A::mData
2          // B::mData1

On a side note, Stanley B. Lippman has excellent book "Inside the C++ Object Model".

Answer 5

One way is to print out the offsets of all the members:

class Parent{
public:
    int a;
    int b;

    virtual void foo(){
        cout << "parent" << endl;
    }
};

class Child : public Parent{
public:
    int c;
    int d;

    virtual void foo(){
        cout << "child" << endl;
    }
};

int main(){

    Parent p;
    Child c;

    p.foo();
    c.foo();

    cout << "Parent Offset a = " << (size_t)&p.a - (size_t)&p << endl;
    cout << "Parent Offset b = " << (size_t)&p.b - (size_t)&p << endl;

    cout << "Child Offset a = " << (size_t)&c.a - (size_t)&c << endl;
    cout << "Child Offset b = " << (size_t)&c.b - (size_t)&c << endl;
    cout << "Child Offset c = " << (size_t)&c.c - (size_t)&c << endl;
    cout << "Child Offset d = " << (size_t)&c.d - (size_t)&c << endl;

    system("pause");
}

Output:

parent
child
Parent Offset a = 8
Parent Offset b = 12
Child Offset a = 8
Child Offset b = 12
Child Offset c = 16
Child Offset d = 20

So you can see all the offsets here. You'll notice that there's nothing at offset 0, as that is presumably where the pointer to the vtable goes.

Also notice that the inherited members have the same offsets in both Child and Parent.